Radical Equations and Extraneous Roots Answers

I do not agree with the student. While solving the equation results in x = 4, this is an extraneous solution. When you substitute 4 back into the original equation, you get sqrt(2(4) + 1) + 3 = 0, which simplifies to 3 + 3 = 0, or 6 = 0. Since this is a false statement, x = 4 is not a solution. Furthermore, the equation sqrt(2x + 1) = -3 has no real solution because a principal square root cannot be negative.